Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $x = \dfrac{a - 5}{-10a - 30} \times \dfrac{4a^2 + 12a}{a^2 - 25} $
Explanation: First factor the quadratic. $x = \dfrac{a - 5}{-10a - 30} \times \dfrac{4a^2 + 12a}{(a - 5)(a + 5)} $ Then factor out any other terms. $x = \dfrac{a - 5}{-10(a + 3)} \times \dfrac{4a(a + 3)}{(a - 5)(a + 5)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (a - 5) \times 4a(a + 3) } { -10(a + 3) \times (a - 5)(a + 5) } $ $x = \dfrac{ 4a(a - 5)(a + 3)}{ -10(a + 3)(a - 5)(a + 5)} $ Notice that $(a + 3)$ and $(a - 5)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 4a\cancel{(a - 5)}(a + 3)}{ -10(a + 3)\cancel{(a - 5)}(a + 5)} $ We are dividing by $a - 5$ , so $a - 5 \neq 0$ Therefore, $a \neq 5$ $x = \dfrac{ 4a\cancel{(a - 5)}\cancel{(a + 3)}}{ -10\cancel{(a + 3)}\cancel{(a - 5)}(a + 5)} $ We are dividing by $a + 3$ , so $a + 3 \neq 0$ Therefore, $a \neq -3$ $x = \dfrac{4a}{-10(a + 5)} $ $x = \dfrac{-2a}{5(a + 5)} ; \space a \neq 5 ; \space a \neq -3 $